Quote:Sacke86Hello world! I want to adapt the classic L'abouchere for 50% winning chances for a 62.5% winning chances strategy to use this system in sports betting with 1.60 medium odds (wich is 62.5% won bets). So how much do I need to raise the bets for 62.5% winrate? For 50% It's simple because we just add the numbers from the ends like this: 10, 10, 20, 30 and we have a bet of 10 + 30 = 40. But with 62.5% winrate and a lower return (x1.6) we must raise the bet more and I don't know how to do it properly to have the same return as this would be a 50% winrate (odds 2) and a return of x2. Sorry for my bad english, this is not my native language! Thanks!

Are you saying you are able to pick winners 62.5% of the time on even money bets? Or that you have to bet $16 to win $10 but will be successful 62.5% of the time? I think you mean the latter?

For example, start with 10, 10, 20, 30. Your first bet is 40.

If it loses, then instead of adding 40 to the end of the list, add 67 (= 40 x 5/3, rounded up) to the list.

The list is now 10, 10, 20, 30, 67; your next bet is 77.

Quote:ThatDonGuyI think what you have to do is, start with a normer Laboucherie, but when you lose, multiply the bet by 5/3 before writing the number down.

For example, start with 10, 10, 20, 30. Your first bet is 40.

If it loses, then instead of adding 40 to the end of the list, add 67 (= 40 x 5/3, rounded up) to the list.

The list is now 10, 10, 20, 30, 67; your next bet is 77.

You god damn right, thanks!

I made some wrong calculations with your numbers and then I've tried with 1.50 odds and found that some number (let's say V) with wich I must multiply the numbers in the series should solve the equation V x (1.50-1) = 1 and was 2 because 2 x (1.50 - 1) = 2 x 0.5 = 1 and this equation should be valid for every odds, then I calculated V = 1 / (1.60-1) = 1 / 0.60 = 1.66. I even tested it and it works because I get the same return when I end the series of numbers, wich is 60% from the initial bet (+$12 in this case because the original bet is $20):

Case 1 - If I win the first bet:

10, 10 : Bet = 10 + 10 = $20 -> Win = $20 x 1.6 = $32 -> Profit = $32 - $20 = +$12

Case 2 - If I loose the first bet, win the second and third one:

10, 10 : Bet = 10 + 10 = $20 -> Loose = $20 -> Total profit: -$20

10, 10, (10 + 10) x 1.66 : Bet = 10 + 20 x 1.66 = 10 + 33.333 = $43.333 ->

->Win = $43.333 x 1.6 = $69.12 -> Total profit: $69.333 - $43.333 - $20 (lost on the first bet) = +$6

10: Bet = $10 -> Win = $10 x 1.6 = 16 -> Total profit: $16 - $10 = +$6 (from this bet) + 6 (from the second bet above) = +12

I think the result will always be +$12, no matter how long the series will go but I didn't tested that. I hope the case 2 it's enough :))